Question

In an electron microscope, accelerated electrons have wavelength of 0.011 nm. Calculate the voltage through which electrons were accelerated to attain this wavelength.

(Take e = 1.6 × 10⁻¹⁹ C, m_e = 9 × 10⁻³¹ kg, h = 6.6 × 10⁻³⁴ J.s)

Answer 1

Given: Planck’s constant (h) = 6.6 × 10⁻³⁴ J.s

Mass of electron (m_e) = 9 × 10⁻³¹ kg

Charge of electron (e) = 1.6 × 10⁻¹⁹ C

Wavelength (λ) = 0.011 nm = 0.011 × 10⁻⁹ m

The de Broglie wavelength (λ) of an electron is related to its momentum (ρ) by the formula:

λ = `h/rho`

The kinetic energy (K) of an electron accelerated through a voltage (V) is K = eV.

Since, K = `rho^2/(2 m_e)`

We can express momentum as:

ρ = `sqrt(2 m_e eV)`

Substituting this into the wavelength formula gives:

λ = `h/(sqrt(2 m_e eV))`

To find the accelerating voltage (V), rearrange the formula:

V = `h^2/(2 m_e e lambda^2)`

= `(6.6 xx 10^-34)^2/(2 xx (9 xx 10^-31) xx (1.6 xx 10^-19) xx (0.011 xx 10^-9)^2)`

= `(43.56 xx 10^-68)/(2 xx 9 xx 16 xx 10^(-31 -19 - 22))`

= `(43.56 xx 10^-68)/(34.848 xx 10^-72)`

= 1.25 × 10⁴ V

= 12500 V

= 12.5 kV